# ordsets

## Functions for manipulating sets as ordered lists.

Sets are collections of elements with no duplicate elements.
An `ordset`

is a representation of a set, where an ordered
list is used to store the elements of the set. An ordered list
is more efficient than an unordered list. Elements are ordered
according to the *Erlang term order*.

This module provides the same interface as the
`sets(3)`

module
but with a defined representation. One difference is
that while `sets`

considers two elements as different if they
do not match (`=:=`

), this module considers two elements as
different if and only if they do not compare equal (`==`

).

#### Functions

### add_element(Element, Ordset1) -> Ordset2

Returns a new ordered set formed from

with

inserted.

### del_element(Element, Ordset1) -> Ordset2

`Element = term()`

`Ordset1 = Ordset2 = ordset(T)`

Returns

, but with

removed.

### filter(Pred, Ordset1) -> Ordset2

`Pred = fun((Element :: T) -> boolean())`

`Ordset1 = Ordset2 = ordset(T)`

Filters elements in

with boolean function

.

### fold(Function, Acc0, Ordset) -> Acc1

`Function =`

fun((Element :: T, AccIn :: term()) -> AccOut :: term())`Ordset = ordset(T)`

`Acc0 = Acc1 = term()`

Folds

over every element in

and returns the final value of the
accumulator.

### from_list(List) -> Ordset

`List = [T]`

`Ordset = ordset(T)`

Returns an ordered set of the elements in

.

### intersection(OrdsetList) -> Ordset

Returns the intersection of the non-empty list of sets.

### intersection(Ordset1, Ordset2) -> Ordset3

`Ordset1 = Ordset2 = Ordset3 = ordset(term())`

Returns the intersection of

and

.

### is_disjoint(Ordset1, Ordset2) -> boolean()

`Ordset1 = Ordset2 = ordset(term())`

Returns `true`

if

and

are disjoint (have no elements in common),
otherwise `false`

.

### is_element(Element, Ordset) -> boolean()

`Element = term()`

`Ordset = ordset(term())`

Returns `true`

if

is an element of

, otherwise `false`

.

### is_set(Ordset) -> boolean()

`Ordset = term()`

Returns `true`

if

is an ordered set
of elements, otherwise `false`

.

### is_subset(Ordset1, Ordset2) -> boolean()

`Ordset1 = Ordset2 = ordset(term())`

Returns `true`

when every element of

is also a member of

, otherwise
`false`

.

### new() -> []

Returns a new empty ordered set.

### subtract(Ordset1, Ordset2) -> Ordset3

`Ordset1 = Ordset2 = Ordset3 = ordset(term())`

Returns only the elements of

that are not
also elements of

.

### union(OrdsetList) -> Ordset

Returns the merged (union) set of the list of sets.